Question #d6cdd

3 Answers
Apr 19, 2017

The correct answer is d. See below.

Explanation:

Make sure you understand how to work with fractions. I would use the same approach as you wished to use.

We use tanx = sinx/cosx and cotx = cosx/sinx:

=1 - sin^2x/(1 + cosx/sinx)- cos^2x/(1 +sinx/cosx)

=1 - sin^2x/((sinx + cosx)/sinx) - cos^2x/((cosx + sinx)/cosx)

=1 - ((sin^2x)sinx)/(sinx + cosx) - (cos^2x(cosx))/(cosx+ sinx)

=(sinx + cosx - sin^3x - cos^3x)/(cosx + sinx)

You can factor the expression in the numerator.

=(sinx(1 - sin^2x) + cosx(1 - cos^2x))/(cosx + sinx)

Use sin^2x + cos^2x = 1.

=(sinx(cos^2x) + cosx(sin^2x))/(cosx + sinx)

=(sinxcosx(cosx + sinx))/(cosx + sinx)

=sinxcosx

So, the answer is d.

Hopefully this helps!

Apr 19, 2017

Refer to the Explanation.

Explanation:

The Expression=1-sin^2x/(1+cotx)-cos^2x/(1+tanx)

=1-sin^2x/{1+cosx/sinx}-cos^2x/{1+sinx/cosx}

=1-sin^3x/(sinx+cosx)-cos^3x/(cosx+sinx)

=1-{(sin^3x+cos^3x)/(sinx+cosx)}

Using, a^3+b^3=(a+b)(a^2-ab+b^2), we get,

"The Exp."=1-{(sinx+cosx)(sin^2x-sinxcosx+cos^2x)}/(sinx+cosx)

=1-{(1-sinxcosx)},.........[because, sin^2x+cos^2x=1]

:." The Exp.="sinxcosx, i.e., "Option d)."

Enjoy Maths.!

Apr 19, 2017

"D" is the Correct Answer

Explanation:

first of all factor the (-1) from -sin^2x/(1+cotx)-cos^2x/(1+tanx)

you get 1-(sin^2x/(1+cotx)+cos^2x/(1+tanx))

simplify each term on its own then add them

sin^2x/(1+cotx) = sin^2/(1+cosx/sinx = sin^2/((sinx+cosx)/sinx) = sin^3x/(sinx+cosx) -->"1"

cos^2x/(1+tanx) = cos^2x/(1+sinx/cosx) = cos^2x/((sinx+cosx)/cosx) = cos^3x/(cosx+sinx) --> "2"

summing 1,2

(sin^3x+cos^3x)/(sinx+cosx)

= ((sinx+cosx)(sin^2x-sinx*cosx+cos^2x))/(sinx+cosx)

"cancelling Sinx+cosx" : (sin^2x-sinx*cosx+cos^2x) =

1-cosx*sinx

back to the full view : 1-(1-Cosx+sinx) = Cosx+sinx